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Mathematics 20 Online
OpenStudy (anonymous):

The times of all 15 year olds who run a certain race are approximately normally distributed with a given mean = 18 sec and standard deviation = 1.2 sec. What percentage of the runners have times less than 14.4 sec? A)0.15% B)0.30% C)0.60% D)2.50%

OpenStudy (kirbykirby):

If \(X\) is the time of a 15-yr old runner, then you are interested in \(P(X < 14.4)\). You can standardize \(X\) with the formula: \(Z=\dfrac{X-\mu}{\sigma} \) where \(\mu\) is the mean, \(\sigma\) is the std. deviation.

OpenStudy (kirbykirby):

\[P(X<14.4)=P\left(\frac{X-18}{1.2} <\frac{14.4-18}{1.2}\right)=P(Z<-3) \]

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